Question

For the circuit given in Problem 13.3, replace the circuit elements by their discrete-time equivalent circuits and write nodal equations in a form suitable for computer solution of the sending-end and receiving-end voltages. Give equations for all dependent sources. Assume, $\mathbf{E}\mathbf{=}\mathbf{1000}\mathbf{V}$, ${\mathbf{L}}_{\mathbf{R}}\mathbf{=}\mathbf{20}\mathbf{mH}$, ${\mathbf{Z}}_{\mathbf{c}}\mathbf{=}\mathbf{100}\mathbf{\Omega }$, $\mathbf{v}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$, $\mathbf{I}\mathbf{=}\mathbf{40}\mathbf{}\mathbf{km}$, and $\mathbf{∆}\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{}\mathbf{ms}$.

Short answer

The nodal equations for ${\mathrm{V}}_{\mathrm{k}}\left(\mathrm{t}\right)$, ${\mathrm{v}}_{\mathrm{m}}\left(\mathrm{t}\right)$, ${\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}\right)$, ${\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}\right)$and ${\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}\right)$are $\left[10-{\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)\right]50\mathrm{V}$, $\left[{\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)-{\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)\right]\times 95.238$, ${\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)-0.02{\mathrm{v}}_{\mathrm{m}}\left(\mathrm{t}\right)\mathrm{A}$, ${\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)+0.02{\mathrm{v}}_{\mathrm{k}}\left(\mathrm{t}\right)\mathrm{A}$, and${\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}-0.02\times {10}^{-3}\right)+0.001{\mathrm{v}}_{\mathrm{m}}\left(\mathrm{t}\right)\mathrm{A}$ respectively.

Step-by-step solution

Write the given data from the question.

The voltage,$\mathrm{E}=1000\mathrm{V}$

Velocity,$\mathrm{v}=2\times {10}^8\mathrm{m}/\mathrm{s}$

Load inductance,${\mathrm{L}}_{\mathrm{R}}=20\mathrm{mH}$

Characteristics impedance,${\mathrm{Z}}_{\mathrm{c}}=100\mathrm{\Omega }$

The length of the line,$\mathrm{l}=40\mathrm{km}$

Time, $∆\mathrm{t}=0.02\mathrm{ms}$

Determine the equation to write the nodal equalisations.

The equation to calculate the transit time is given as follows.

$\mathbf{\tau }\mathbf{=}\frac{\mathbf{1}}{\mathbf{\nu }}$ …… (1)

Write the nodal equation.

Consider the circuit diagram using the terminal variable in the domain.

Consider the circuit diagram,

Calculate the transit time.

Substitute$40\mathrm{km}$ for$\mathrm{l}$ and$2\times {10}^8\mathrm{m}/\mathrm{s}$ into equation (1).

$\begin{array}{rcl}\mathrm{r}& =& \frac{40\times {10}^3}{2\times {10}^8}\\ & =& 20\times {10}^{-5}\\ & =& 0.2\mathrm{ms}\end{array}$

Apply the nodal analysis at node 1.

${\mathrm{V}}_{\mathrm{k}}\left(\mathrm{t}\right)=\left[{\mathrm{i}}_{\mathrm{G}}\left(\mathrm{t}\right)-{\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}-\mathrm{\zeta }\right)\right]\left({\mathrm{Z}}_{\mathrm{c}}{\mathrm{PZ}}_{\mathrm{G}}\right)$

Substituterole="math" localid="1656675846985" $\frac{{\mathrm{e}}_{\mathrm{G}}\left(\mathrm{t}\right)}{{\mathrm{Z}}_{\mathrm{G}}}$ for${\mathrm{i}}_{\mathrm{G}}$ into above equation.

${\mathrm{V}}_{\mathrm{k}}\left(\mathrm{t}\right)=\left[\frac{{\mathrm{e}}_{\mathrm{G}}\left(\mathrm{t}\right)}{{\mathrm{Z}}_{\mathrm{G}}}-{\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}-\mathrm{\zeta }\right)\right]\left({\mathrm{Z}}_{\mathrm{c}}{\mathrm{PZ}}_{\mathrm{G}}\right)$

Substitute $0.2\mathrm{ms}$for $\zeta$, $1000\mathrm{V}$for ${\mathrm{e}}_{\mathrm{G}}\left(\mathrm{t}\right)$, $100\mathrm{\Omega }$for ${\mathrm{Z}}_{\mathrm{c}}$and data-custom-editor="chemistry" ${\mathrm{Z}}_{\mathrm{G}}$into above equation.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{k}}\left(\mathrm{t}\right)=\left[\frac{1000}{100}-{\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)\right]\left(100\mathrm{P}100\right) \\ {\mathrm{V}}_{\mathrm{k}}\left(\mathrm{t}\right)=\left[10-{\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)\right]50\mathrm{V} \end{gathered}$$

Apply the nodal analysis at node 2,

role="math" localid="1656677099528" ${\mathrm{V}}_{\mathrm{m}}\left(\mathrm{t}\right)=\left[{\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}-\mathrm{\zeta }\right)-{\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}-∆\mathrm{t}\right)\right]\left({\mathrm{Z}}_{\mathrm{c}}\mathrm{P}\frac{2\mathrm{L}}{∆\mathrm{t}}\right)$

Substitute $0.2\mathrm{ms}$for $\zeta$, $0.02\mathrm{ms}$for $∆\mathrm{t}$, $100\mathrm{\Omega }$for ${\mathrm{Z}}_{\mathrm{c}}$and $20\mathrm{mH}$for $\mathrm{L}$into above equation.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{m}}\left(\mathrm{t}\right)=\left[{\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)-{\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}-0.02\times {10}^{-3}\right)\right]\left(100\mathrm{P}\frac{2\left(20\times {10}^{-3}\right)}{0.02\times {10}^{-3}}\right) \\ {\mathrm{V}}_{\mathrm{m}}\left(\mathrm{t}\right)=\left[{\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)-{\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}-0.02\times {10}^{-3}\right)\right]\left(100\mathrm{P}2000\right) \\ {\mathrm{V}}_{\mathrm{m}}\left(\mathrm{t}\right)=\left[{\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)-{\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}-0.02\times {10}^{-3}\right)\right]\times 95.238 \end{gathered}$$

The expression for the current${\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}-\mathrm{\zeta }\right)$ .

${\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}-\mathrm{\zeta }\right)={\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}-2\mathrm{\zeta }\right)-\frac{2}{{\mathrm{Z}}_{\mathrm{c}}}{\mathrm{v}}_{\mathrm{m}}\left(\mathrm{t}-\mathrm{\zeta }\right)$

Replace$\mathrm{t}-\mathrm{\zeta }$ with$\mathrm{t}$ into above equation.

${\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}\right)={\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}-\mathrm{\zeta }\right)-\frac{2}{{\mathrm{Z}}_{\mathrm{c}}}{\mathrm{v}}_{\mathrm{m}}\left(\mathrm{t}\right)$

Substitute$0.2\mathrm{ms}$ for$\zeta$ and$100\mathrm{\Omega }$ for${\mathrm{Z}}_{\mathrm{c}}$ into above equation.

$$\begin{gathered} {\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}\right)={\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)-\frac{2}{100}{\mathrm{v}}_{\mathrm{m}}\left(\mathrm{t}\right) \\ {\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}\right)={\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)-0.02{\mathrm{v}}_{\mathrm{m}}\left(\mathrm{t}\right) \end{gathered}$$

The expression for the current${\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}-\mathrm{\zeta }\right)$ .

${\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}-\mathrm{\zeta }\right)={\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}-2\mathrm{\zeta }\right)+\frac{2}{{\mathrm{Z}}_{\mathrm{c}}}{\mathrm{v}}_{\mathrm{k}}\left(\mathrm{t}-\mathrm{\zeta }\right)$

Replace$\mathrm{t}-\mathrm{\zeta }$ with$\mathrm{t}$ into above equation.

${\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}\right)={\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}-\mathrm{\zeta }\right)+\frac{2}{{\mathrm{Z}}_{\mathrm{c}}}{\mathrm{v}}_{\mathrm{k}}\left(\mathrm{t}\right)$

Substitute$0.2\mathrm{ms}$ for$\zeta$ and$100\mathrm{\Omega }$ for${\mathrm{Z}}_{\mathrm{c}}$ into above equation.

$$\begin{gathered} {\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}\right)={\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)+\frac{2}{100}{\mathrm{v}}_{\mathrm{k}}\left(\mathrm{t}\right) \\ {\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}\right)={\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)+0.02{\mathrm{v}}_{\mathrm{k}}\left(\mathrm{t}\right)\mathrm{A} \end{gathered}$$

The expression for the current${\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}-∆\mathrm{t}\right)$ .

${\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}-∆\mathrm{t}\right)={\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}-2∆\mathrm{t}\right)+\frac{{\mathrm{v}}_{\mathrm{m}}\left(\mathrm{t}-∆\mathrm{t}\right)}{{\displaystyle \raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$∆\mathrm{t}$}\right.}}$

Replace$\mathrm{t}-∆\mathrm{t}$ with$\mathrm{t}$ into above equation.

${\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}\right)={\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}-∆\mathrm{t}\right)+\frac{{\mathrm{v}}_{\mathrm{m}}\left(\mathrm{t}\right)}{{\displaystyle \raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$∆\mathrm{t}$}\right.}}$

Substitute $0.02\mathrm{ms}$for $∆\mathrm{t}$and $20\mathrm{ms}$for $\mathrm{L}$into above equation.

$$\begin{gathered} {\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}\right)={\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}-0.02\times {10}^{-3}\right)+\frac{{\mathrm{v}}_{\mathrm{m}}\left(\mathrm{t}\right)}{{\displaystyle \frac{20\times {10}^{-3}}{0.02\times {10}^{-3}}}} \\ {\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}\right)={\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}-0.02\times {10}^{-3}\right)+\frac{{\mathrm{v}}_{\mathrm{m}}\left(\mathrm{t}\right)}{{\displaystyle 1000}} \\ {\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}\right)={\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}-0.02\times {10}^{-3}\right)+0.001{\mathrm{v}}_{\mathrm{m}}\left(\mathrm{t}\right)\mathrm{A} \end{gathered}$$

Hence the nodal equations for V_k(t), v_m(t), I_k(t), I_m(t) and I_L(t) are $\left[10-{\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)\right]50\mathrm{V}$, $\left[{\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)-{\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}-0.02\times {10}^{-3}\right)\right]\times 95.238$, ${\mathrm{I}}_{\mathrm{m}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)-0.02{\mathrm{v}}_{\mathrm{m}}\left(\mathrm{t}\right)\mathrm{A}$, ${\mathrm{I}}_{\mathrm{k}}\left(\mathrm{t}-0.2\times {10}^{-3}\right)+0.02{\mathrm{v}}_{\mathrm{k}}\left(\mathrm{t}\right)\mathrm{A}$and ${\mathrm{I}}_{\mathrm{L}}\left(\mathrm{t}-0.02\times {10}^{-3}\right)+0.001{\mathrm{v}}_{\mathrm{m}}\left(\mathrm{t}\right)\mathrm{A}$respectively.